{"id":3768,"date":"2021-05-12T20:24:06","date_gmt":"2021-05-12T20:24:06","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/1d\/"},"modified":"2021-05-12T20:24:06","modified_gmt":"2021-05-12T20:24:06","slug":"1d","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/1d\/","title":{"raw":"Skills Review for Inverse Functions","rendered":"Skills Review for Inverse Functions"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Solve rational equations by clearing denominators<\/li>\n \t<li>Solve for a variable in a square root equation<\/li>\n \t<li>Convert between radical and exponent notations<\/li>\n \t<li>Solve for a variable in a complex radical or rational exponent equation<\/li>\n<\/ul>\n<\/div>\nThe Inverse Functions section covers in great detail everything you need to know about inverse functions. Depending on the type of function you are working with, finding the inverse of a function algebraically can require you to manipulate and rearrange a function's equation quite a bit. Here we will review some of the techniques that can be used to rearrange certain types of equations and solve for a specified variable.\n<h2>Manipulate Rational Equations<\/h2>\nEquations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex] \\frac{2y+1}{4}=\\frac{x}{3}[\/latex] is a rational equation (of two variables).\n\nOne of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.\n\nSolve for [latex]y[\/latex] in the equation [latex]\\frac{1}{2}y-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.\n\nMultiply both sides of the equation by&nbsp;[latex]4[\/latex], the common denominator of the fractional coefficients.\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{1}{2}y-3=2-\\frac{3}{4}x\\\\ 4\\left(\\frac{1}{2}y-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\\\\\text{}\\,\\,\\,\\,4\\left(\\frac{1}{2}y\\right)-4\\left(3\\right)=4\\left(2\\right)+4\\left(-\\frac{3}{4}x\\right)\\\\2y-12=8-3x\\\\\\underline{+12}\\,\\,\\,\\,\\,\\,\\underline{+12}\\\\ 2y=20-3x\\\\ \\frac{2y}{2}=\\frac{20-3x}{2} \\\\ y=10-\\frac{3}{2}x\\end{array}[\/latex]<\/p>\n\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\nSolve the equation [latex] \\frac{x+5}{8}=\\frac{7}{y}[\/latex] for [latex]y[\/latex].\n[reveal-answer q=\"425621\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"425621\"]\n\nFind the least common denominator of&nbsp;[latex]8[\/latex] and&nbsp;[latex]y[\/latex]. [latex]8y[\/latex] will be the LCD.\n\nMultiply both sides of the equation by the common denominator,&nbsp;[latex]8y[\/latex], to keep the equation balanced and to eliminate the denominators.\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8y\\cdot \\frac{x+5}{8}=\\frac{7}{y}\\cdot 8y\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y(x+5)}{8}=\\frac{7(8y)}{y}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y}{8}\\cdot (x+5)=\\frac{7(8y)}{y}\\\\\\\\\\frac{8}{8}\\cdot y(x+5)=7\\cdot 8\\cdot \\frac{y}{y}\\\\\\\\1\\cdot y(x+5)=56\\cdot 1\\,\\,\\,\\\\\\\\ y(x+5)=56\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ \\frac{y(x+5)}{x+5}=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ y=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\nIn the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\nSolve the equation [latex] x=\\frac{4}{3y+1}[\/latex] for [latex]y[\/latex].\n\n[reveal-answer q=\"331190\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"331190\"]\n\nClear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3y+1[\/latex].\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{3y+1}\\right)\\left(3y+1\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Simplify common factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{\\cancel{3y+1}}\\right)\\left(\\cancel{3y+1}\\right)\\\\\\left(3y+1\\right)\\left(x\\right)=4\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We can now use the addition and multiplication properties of equality to solve it.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=4 \\\\ \\\\ \\frac{(3y+1)(x)}{x}=\\frac{4}{x} \\\\ \\\\ 3y+1=\\frac{4}{x} \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\,\\,\\,\\underline{-1} \\\\ 3y=\\frac{4}{x}-1\\\\ \\\\ \\frac{3y}{3}=\\frac{4}{3x}-\\frac{1}{3}\\\\ \\\\ y=\\frac{4}{3}x-\\frac{1}{3}\\end{array}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<h2>Manipulate Radical Equations<\/h2>\n<strong>Radical equations<\/strong> are equations that contain variables in the <strong>radicand<\/strong> (the expression under a radical symbol), such as\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\sqrt{3y+18}=x &amp; \\\\ \\sqrt{x+3}=y-3 &amp; \\\\ \\sqrt{x+5}-\\sqrt{y - 3}=2\\end{array}[\/latex]<\/div>\nRadical equations are manipulated by eliminating each radical, one at a time until you have solved for the indicated variable.\n<div class=\"textbox\">\n<h3>A General Note: Radical Equations<\/h3>\nAn equation containing terms with a variable in the radicand is called a <strong>radical equation<\/strong>.\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a radical equation, solve it<\/h3>\n<ol>\n \t<li>Isolate the radical expression containing your variable of interest on one side of the equal sign. Put all remaining terms on the other side.<\/li>\n \t<li>If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an <em>n<\/em>th root radical, raise both sides to the <em>n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\n \t<li>Solve the resulting equation for the variable of interest.<\/li>\n \t<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation with One Radical<\/h3>\nSolve [latex]\\sqrt{15 - 2y}=x[\/latex] for [latex]y[\/latex].\n\n[reveal-answer q=\"503795\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"503795\"]\n\nThe radical is already isolated on the left side of the equal sign, so proceed to square both sides.\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\sqrt{15 - 2y}=x &amp; \\\\ {\\left(\\sqrt{15 - 2y}\\right)}^{2}={\\left(x\\right)}^{2} &amp; \\\\ 15 - 2y={x}^{2}\\end{array}[\/latex]<\/div>\nNow isolate [latex]y[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll} 15 - 2y={x}^{2} \\\\ -2y={x}^{2}-15 \\\\ y=-\\frac{1}{2}{x}^{2}+\\frac{15}{2}\\end{array}[\/latex]<\/div>\n[\/hidden-answer]\n\n&nbsp;\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nSolve the radical equation: [latex]\\sqrt{y+3}=3x - 1[\/latex]\n\n[reveal-answer q=\"719648\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"719648\"]\n\n[latex]y=9{x}^{2}-6x-2[\/latex] Note: When you square the binomial on the right side of the equation you get [latex]9{x}^{2}-6x+1[\/latex]. Then, take away [latex]3[\/latex] from both sides.[\/hidden-answer]\n\n<\/div>\nRational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\\frac{1}{2}}[\/latex] is another way of writing [latex]\\sqrt{16}[\/latex] and [latex]{8}^{\\frac{2}{3}}[\/latex] is another way of writing [latex]\\left(\\sqrt[3]{8}\\right)^2[\/latex].\n<p style=\"padding-left: 30px;\">We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\\frac{2}{3}\\left(\\frac{3}{2}\\right)=1[\/latex].<\/p>\n\n<div class=\"textbox\">\n<h3>A General Note: Rational Exponents<\/h3>\nA rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:\n<div style=\"text-align: center;\">[latex]{a}^{\\frac{m}{n}}={\\left({a}^{\\frac{1}{n}}\\right)}^{m}={\\left({a}^{m}\\right)}^{\\frac{1}{n}}=\\sqrt[n]{{a}^{m}}={\\left(\\sqrt[n]{a}\\right)}^{m}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation involving a Variable raised to a Rational Exponent<\/h3>\nSolve the equation in which a variable is raised to a rational exponent: [latex]{x}^{\\frac{5}{4}}=32[\/latex].\n\n[reveal-answer q=\"887306\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"887306\"]\n\nThe way to remove the exponent on <em>x<\/em> is by raising both sides of the equation to a power that is the reciprocal of [latex]\\frac{5}{4}[\/latex], which is [latex]\\frac{4}{5}[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllll}{x}^{\\frac{5}{4}}=32\\hfill &amp; \\hfill &amp; \\\\ {\\left({x}^{\\frac{5}{4}}\\right)}^{\\frac{4}{5}}={\\left(32\\right)}^{\\frac{4}{5}}\\hfill &amp; \\hfill &amp; \\\\ x={\\left(2\\right)}^{4}\\hfill &amp; \\text{The fifth root of 32 is 2}.\\hfill &amp; \\\\ x=16\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\n<div>[\/hidden-answer]<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nSolve the equation [latex]{x}^{\\frac{3}{2}}=125[\/latex].\n\n[reveal-answer q=\"390459\"]Show Solution[\/reveal-answer][hidden-answer a=\"390459\"]\n\n[latex]25[\/latex][\/hidden-answer]\n\n[ohm_question]38391[\/ohm_question]\n\n<\/div>\nNote: In the two examples above, there was only one variable in the equations. So, we solved for the variable and found a numeric solution. In the case where there are two different variables in the equation and you are asked to solve for one in terms of the other, you still follow the same process as you solve for the indicated variable.\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve rational equations by clearing denominators<\/li>\n<li>Solve for a variable in a square root equation<\/li>\n<li>Convert between radical and exponent notations<\/li>\n<li>Solve for a variable in a complex radical or rational exponent equation<\/li>\n<\/ul>\n<\/div>\n<p>The Inverse Functions section covers in great detail everything you need to know about inverse functions. Depending on the type of function you are working with, finding the inverse of a function algebraically can require you to manipulate and rearrange a function&#8217;s equation quite a bit. Here we will review some of the techniques that can be used to rearrange certain types of equations and solve for a specified variable.<\/p>\n<h2>Manipulate Rational Equations<\/h2>\n<p>Equations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex]\\frac{2y+1}{4}=\\frac{x}{3}[\/latex] is a rational equation (of two variables).<\/p>\n<p>One of the most straightforward ways to solve a rational equation for the indicated variable is to eliminate denominators with the common denominator and then use properties of equality to isolate the indicated variable.<\/p>\n<p>Solve for [latex]y[\/latex] in the equation [latex]\\frac{1}{2}y-3=2-\\frac{3}{4}x[\/latex] by clearing the fractions in the equation first.<\/p>\n<p>Multiply both sides of the equation by&nbsp;[latex]4[\/latex], the common denominator of the fractional coefficients.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\frac{1}{2}y-3=2-\\frac{3}{4}x\\\\ 4\\left(\\frac{1}{2}y-3\\right)=4\\left(2-\\frac{3}{4}x\\right)\\\\\\text{}\\,\\,\\,\\,4\\left(\\frac{1}{2}y\\right)-4\\left(3\\right)=4\\left(2\\right)+4\\left(-\\frac{3}{4}x\\right)\\\\2y-12=8-3x\\\\\\underline{+12}\\,\\,\\,\\,\\,\\,\\underline{+12}\\\\ 2y=20-3x\\\\ \\frac{2y}{2}=\\frac{20-3x}{2} \\\\ y=10-\\frac{3}{2}x\\end{array}[\/latex]<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve the equation [latex]\\frac{x+5}{8}=\\frac{7}{y}[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q425621\">Show Solution<\/span><\/p>\n<div id=\"q425621\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the least common denominator of&nbsp;[latex]8[\/latex] and&nbsp;[latex]y[\/latex]. [latex]8y[\/latex] will be the LCD.<\/p>\n<p>Multiply both sides of the equation by the common denominator,&nbsp;[latex]8y[\/latex], to keep the equation balanced and to eliminate the denominators.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}8y\\cdot \\frac{x+5}{8}=\\frac{7}{y}\\cdot 8y\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y(x+5)}{8}=\\frac{7(8y)}{y}\\,\\,\\,\\,\\,\\,\\\\\\\\\\frac{8y}{8}\\cdot (x+5)=\\frac{7(8y)}{y}\\\\\\\\\\frac{8}{8}\\cdot y(x+5)=7\\cdot 8\\cdot \\frac{y}{y}\\\\\\\\1\\cdot y(x+5)=56\\cdot 1\\,\\,\\,\\\\\\\\ y(x+5)=56\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ \\frac{y(x+5)}{x+5}=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\\\ y=\\frac{56}{x+5}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they do not share any common factors.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve the equation [latex]x=\\frac{4}{3y+1}[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q331190\">Show Solution<\/span><\/p>\n<div id=\"q331190\" class=\"hidden-answer\" style=\"display: none\">\n<p>Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3y+1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{3y+1}\\right)\\left(3y+1\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Simplify common factors.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=\\left(\\frac{4}{\\cancel{3y+1}}\\right)\\left(\\cancel{3y+1}\\right)\\\\\\left(3y+1\\right)\\left(x\\right)=4\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">We can now use the addition and multiplication properties of equality to solve it.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3y+1\\right)\\left(x\\right)=4 \\\\ \\\\ \\frac{(3y+1)(x)}{x}=\\frac{4}{x} \\\\ \\\\ 3y+1=\\frac{4}{x} \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-1}\\,\\,\\,\\,\\,\\,\\underline{-1} \\\\ 3y=\\frac{4}{x}-1\\\\ \\\\ \\frac{3y}{3}=\\frac{4}{3x}-\\frac{1}{3}\\\\ \\\\ y=\\frac{4}{3}x-\\frac{1}{3}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Manipulate Radical Equations<\/h2>\n<p><strong>Radical equations<\/strong> are equations that contain variables in the <strong>radicand<\/strong> (the expression under a radical symbol), such as<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\sqrt{3y+18}=x & \\\\ \\sqrt{x+3}=y-3 & \\\\ \\sqrt{x+5}-\\sqrt{y - 3}=2\\end{array}[\/latex]<\/div>\n<p>Radical equations are manipulated by eliminating each radical, one at a time until you have solved for the indicated variable.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Radical Equations<\/h3>\n<p>An equation containing terms with a variable in the radicand is called a <strong>radical equation<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a radical equation, solve it<\/h3>\n<ol>\n<li>Isolate the radical expression containing your variable of interest on one side of the equal sign. Put all remaining terms on the other side.<\/li>\n<li>If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an <em>n<\/em>th root radical, raise both sides to the <em>n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\n<li>Solve the resulting equation for the variable of interest.<\/li>\n<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation with One Radical<\/h3>\n<p>Solve [latex]\\sqrt{15 - 2y}=x[\/latex] for [latex]y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q503795\">Show Solution<\/span><\/p>\n<div id=\"q503795\" class=\"hidden-answer\" style=\"display: none\">\n<p>The radical is already isolated on the left side of the equal sign, so proceed to square both sides.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\sqrt{15 - 2y}=x & \\\\ {\\left(\\sqrt{15 - 2y}\\right)}^{2}={\\left(x\\right)}^{2} & \\\\ 15 - 2y={x}^{2}\\end{array}[\/latex]<\/div>\n<p>Now isolate [latex]y[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll} 15 - 2y={x}^{2} \\\\ -2y={x}^{2}-15 \\\\ y=-\\frac{1}{2}{x}^{2}+\\frac{15}{2}\\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the radical equation: [latex]\\sqrt{y+3}=3x - 1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q719648\">Show Solution<\/span><\/p>\n<div id=\"q719648\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=9{x}^{2}-6x-2[\/latex] Note: When you square the binomial on the right side of the equation you get [latex]9{x}^{2}-6x+1[\/latex]. Then, take away [latex]3[\/latex] from both sides.<\/p><\/div>\n<\/div>\n<\/div>\n<p>Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\\frac{1}{2}}[\/latex] is another way of writing [latex]\\sqrt{16}[\/latex] and [latex]{8}^{\\frac{2}{3}}[\/latex] is another way of writing [latex]\\left(\\sqrt[3]{8}\\right)^2[\/latex].<\/p>\n<p style=\"padding-left: 30px;\">We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\\frac{2}{3}\\left(\\frac{3}{2}\\right)=1[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Rational Exponents<\/h3>\n<p>A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{\\frac{m}{n}}={\\left({a}^{\\frac{1}{n}}\\right)}^{m}={\\left({a}^{m}\\right)}^{\\frac{1}{n}}=\\sqrt[n]{{a}^{m}}={\\left(\\sqrt[n]{a}\\right)}^{m}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation involving a Variable raised to a Rational Exponent<\/h3>\n<p>Solve the equation in which a variable is raised to a rational exponent: [latex]{x}^{\\frac{5}{4}}=32[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q887306\">Show Solution<\/span><\/p>\n<div id=\"q887306\" class=\"hidden-answer\" style=\"display: none\">\n<p>The way to remove the exponent on <em>x<\/em> is by raising both sides of the equation to a power that is the reciprocal of [latex]\\frac{5}{4}[\/latex], which is [latex]\\frac{4}{5}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllll}{x}^{\\frac{5}{4}}=32\\hfill & \\hfill & \\\\ {\\left({x}^{\\frac{5}{4}}\\right)}^{\\frac{4}{5}}={\\left(32\\right)}^{\\frac{4}{5}}\\hfill & \\hfill & \\\\ x={\\left(2\\right)}^{4}\\hfill & \\text{The fifth root of 32 is 2}.\\hfill & \\\\ x=16\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the equation [latex]{x}^{\\frac{3}{2}}=125[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q390459\">Show Solution<\/span><\/p>\n<div id=\"q390459\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]25[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm38391\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38391&theme=oea&iframe_resize_id=ohm38391&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Note: In the two examples above, there was only one variable in the equations. So, we solved for the variable and found a numeric solution. In the case where there are two different variables in the equation and you are asked to solve for one in terms of the other, you still follow the same process as you solve for the indicated variable.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3768\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/precalculus\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3768","chapter","type-chapter","status-publish","hentry"],"part":3764,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3768","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3768\/revisions"}],"predecessor-version":[{"id":4994,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3768\/revisions\/4994"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/3764"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3768\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3768"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3768"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3768"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3768"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}